Important Derivations
Step-by-step derivations for Class 12 Physics and Maths. Each one includes the final result, number of steps, marks weightage, and how often it appears in board exams.
Physics (10 derivations)
Electric Charges and Fields
Expression for Electric Field on Axial Line of Dipole
RESULT
E_axial = 2kp/r³ (for r >> 2l)
Steps
- 1
Consider a dipole with charges +q and −q separated by distance 2l
- 2
Take a point P on the axial line at distance r from centre
- 3
E due to +q: E₁ = kq/(r−l)² (towards P)
- 4
E due to −q: E₂ = kq/(r+l)² (away from P)
- 5
E_net = E₁ − E₂ = kq[1/(r−l)² − 1/(r+l)²]
- 6
Simplify and apply r >> l: E_axial = 2kp/r³
Draw the dipole diagram first. Direction of E_axial is along dipole moment direction.
Electric Charges and Fields
Expression for Electric Field on Equatorial Line of Dipole
RESULT
E_equatorial = kp/(r² + l²)^(3/2) ≈ kp/r³ (for r >> l)
Steps
- 1
Take point P on equatorial line at distance r from centre
- 2
E₁ = E₂ = kq/(r² + l²) — magnitudes are equal
- 3
Vertical components cancel; horizontal components add up
- 4
E_net = 2E₁ cosθ where cosθ = l/√(r²+l²)
- 5
Substitute and simplify: E_eq = kp/(r²+l²)^(3/2)
Direction is OPPOSITE to dipole moment — anti-parallel.
Electrostatic Potential and Capacitance
Derivation of Capacitance of Parallel Plate Capacitor
RESULT
C = ε₀A/d
Steps
- 1
Two large parallel plates of area A, separated by distance d, with charges +Q and −Q
- 2
Surface charge density σ = Q/A
- 3
Electric field between plates: E = σ/ε₀ = Q/(ε₀A) (using Gauss's law)
- 4
Potential difference: V = E × d = Qd/(ε₀A)
- 5
Capacitance C = Q/V = ε₀A/d
With dielectric of constant K: C = Kε₀A/d = KC₀.
Moving Charges and Magnetism
Biot-Savart Law — Magnetic Field at Centre of Circular Current Loop
RESULT
B = μ₀I/2R
Steps
- 1
Each element dl of the circular loop is perpendicular to r
- 2
dB = (μ₀/4π) × Idl/R² (since sinθ = 1 for every element)
- 3
All dB elements point in same direction (into or out of loop plane)
- 4
Integrate: B = (μ₀/4π) × I/R² × ∮dl = (μ₀/4π) × I × 2πR/R²
- 5
B = μ₀I/2R
For N turns: B = μ₀NI/2R.
Moving Charges and Magnetism
Ampere's Law — Magnetic Field Inside a Solenoid
RESULT
B = μ₀nI (n = turns per unit length)
Steps
- 1
Consider a rectangular Amperian loop ABCDA of length l inside solenoid
- 2
Only side AB (inside solenoid) contributes to ∮B·dl
- 3
∮B·dl = Bl (contributions from BC, CD, DA are zero)
- 4
Total enclosed current = nIl (n turns per unit length, l = length)
- 5
Applying Ampere's Law: Bl = μ₀nIl → B = μ₀nI
This derivation almost always appears in section D of board exam.
Electromagnetic Induction
Expression for Motional EMF (Rod moving in Magnetic Field)
RESULT
ε = Blv
Steps
- 1
A rod of length l moves with velocity v perpendicular to magnetic field B
- 2
Free electrons experience force: F = qv × B (upward for + charges)
- 3
Electrons accumulate at one end, creating potential difference
- 4
Equilibrium: electric force qE = magnetic force qvB → E = vB
- 5
EMF = E × l = Blv
Electromagnetic Induction
Expression for Energy Stored in an Inductor
RESULT
U = ½LI²
Steps
- 1
Work done against back-EMF when current increases by dI: dW = εI dt
- 2
Back-EMF: ε = L(dI/dt)
- 3
dW = L(dI/dt) × I × dt = LI dI
- 4
Total work (energy stored): U = ∫₀ᴵ LI dI = ½LI²
Ray Optics
Mirror Formula Derivation (Concave Mirror)
RESULT
1/f = 1/v + 1/u
Steps
- 1
Consider object AB beyond C (centre of curvature) of concave mirror
- 2
Two rays from tip B: parallel to principal axis (reflects through F), and through C
- 3
Image A'B' formed between F and C
- 4
Using similar triangles: A'B'/AB = A'P/AP ...(i)
- 5
Also: A'B'/AB = A'F/FP ...(ii)
- 6
From (i) and (ii): A'F/FP = A'P/AP
- 7
Substituting with sign convention and simplifying: 1/v + 1/u = 1/f
Draw the ray diagram clearly — 1 mark for diagram in board exam.
Ray Optics
Lens Maker's Formula
RESULT
1/f = (n−1)[1/R₁ − 1/R₂]
Steps
- 1
Apply refraction formula at first surface (radius R₁): n₂/v₁ − n₁/u = (n₂−n₁)/R₁
- 2
Image I₁ acts as virtual object for second surface
- 3
Apply refraction at second surface (radius R₂): n₁/v − n₂/v₁ = (n₁−n₂)/R₂
- 4
Add both equations (v₁ cancels): n₁/v − n₁/u = (n₂−n₁)[1/R₁ − 1/R₂]
- 5
Divide by n₁: 1/v − 1/u = (n−1)[1/R₁ − 1/R₂] = 1/f
Atoms
Bohr's Model — Radius of nth Orbit
RESULT
rₙ = n²a₀/Z where a₀ = 0.529 Å
Steps
- 1
Centripetal force = Coulomb force: mv²/r = Ze²/4πε₀r²
- 2
Bohr's quantisation condition: mvr = nh/2π
- 3
From quantisation: v = nh/2πmr, substitute in force equation
- 4
Solve for r: r = n²h²ε₀/πme²Z = n²(0.529Å)/Z
Also derive energy of nth orbit (E_n = −13.6 Z²/n² eV) from same steps.
Mathematics (3 proofs)
Inverse Trigonometric Functions
Proof: tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) for xy < 1
RESULT
tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)]
Steps
- 1
Let tan⁻¹x = α and tan⁻¹y = β, so tanα = x and tanβ = y
- 2
tan(α+β) = (tanα + tanβ)/(1 − tanα tanβ) = (x+y)/(1−xy)
- 3
Therefore α + β = tan⁻¹[(x+y)/(1−xy)]
- 4
i.e., tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] when xy < 1
Integrals
Integration by Parts Proof (ILATE rule basis)
RESULT
∫u dv = uv − ∫v du
Steps
- 1
Start from product rule: d/dx(uv) = u(dv/dx) + v(du/dx)
- 2
Integrate both sides: uv = ∫u(dv/dx)dx + ∫v(du/dx)dx
- 3
Rearrange: ∫u(dv/dx)dx = uv − ∫v(du/dx)dx
- 4
In shorthand: ∫u dv = uv − ∫v du
ILATE helps choose u: Inverse trig > Log > Algebraic > Trig > Exponential.
Continuity and Differentiability
Rolle's Theorem — Statement and Geometric Interpretation
RESULT
If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then ∃ c ∈ (a,b) such that f'(c) = 0
Steps
- 1
Statement: state the three conditions precisely
- 2
Geometric meaning: if a smooth curve has equal values at endpoints, the tangent is horizontal at some intermediate point
- 3
Proof idea: By extreme value theorem, f has max M and min m on [a,b]
- 4
If M = m, f is constant ⟹ f'(c) = 0 for all c
- 5
If M ≠ m, at least one extreme occurs at interior point c ⟹ f'(c) = 0